How do you calculate the equilibrium concentration of #H_3O+# in a 0.20 M solution of oxalic acid?

Question

Luna Chen · Accepted Answer

#["H"_3"O"^(+)] = 0.0803 color(white)(l) "mol" * "dm"^(-3)#

The oxalic acid is a dicarboxylic acid. Each molecule of the acid partially disassociates to produce up to two hydronium ions when dissolved in water. Calculation of #"H"_3"O"^+# for the #0.20 color(white)(l) "M"# solution requires knowledge of both first and the second acid disassociation constants of the oxalic acid.

From the oxalic acid page on Wikipedia,

Construct a rice table (also in #"M"#) for each of the process. Let the final increase in hydronium concentration due to the disassociation of #"C"_2"H"_2"O"_4# be #x#.

R #"C"_2"H"_2"O"_4 (aq) + "H"_2"O"(l) rightleftharpoons "C"_2"H""O"_4^(-)(aq) + "H"_3"O"^(+)(aq)# I #color(white)(I)0.20 color(white)(."O"_4 (aq) + "H"_2"O"(l) rightleftharpoons) 0 color(white)("."_2"H""O"_4^(-)(aq) + ) 0 # C #-x color(white)("l"_2"O"_4 (aq) + "H"_2"O"(l)rightleftharpoons) +x color(white)("O"_4^(-)(aq) + ) +x# E #0.20 - x color(white)((aq) + "H"_2"O"(l) rightleftharpoons) color(grey)(cancel(x)) color(white)("."_2"H""O"_4^(-)(aq) + ) color(grey)(cancel(x))#

The disassociation of #"C"_2"H"_2"O"_4# accounts for all #"H"_3"O"^(+)# and #"C"_2"H" "O"_4^(-)# in the system before the onset of the second process. Concentrations for these two species in the initial conditions of the second RICE table for the disassociation of #"C"_2"H" "O"_4^(-)# are therefore identical to those in the final condition of the first process. Let the increase in hydronium concentration due to the disassociation of #"C"_2"H" "O"_4^(-)# be #y#.

R #"C"_2"H""O"_4^(-) (aq) + "H"_2"O"(l) rightleftharpoons "C"_2"O"_4^(2-)(aq) + "H"_3"O"^(+)(aq)# I #color(white)(I)x color(white)("c"_2"H""O"_4 (aq) + "H"_2"O"(l) rightleftharpoons) 0 color(white)("."_2"O"_4^(2-)(aq) + ) x # C #-y color(white)("l"_2"O"_4 (aq) + "H"_2"O"(l)rightleftharpoons) +y color(white)("."_4^(2-)(aq) + ) +y# E # color(purple)(x - y) color(white)("C"_4 (aq) + "H"_2"O"(l) rightleftharpoons) y color(white)("."_2"l""O"_4^(-)(aq) + ) color(purple)(x + y)#

The solution to the two RICE table shall satisfy the equilibrium condition for both processes simultaneously. However, both equations are supposed to take the equilibrium concentration found in the second and the last table in case the concentration of a species (e.g., #"H"_3"O"^(+)# is changed in both processes. That is:

Solving the quadratic system (preferably with a Computer Algebra System) yields a collection of sets of values for #x# and #y#. Both #x# and #y# are supposed to be greater than zero given that concentrations of species such as #"H"_3"O"^(+)# and #"C"_2"O"_4^(-)# always take non-negative values. Screening the result with the condition #x >= 0#, #y >= 0# help narrows down to a single possible outcome:

The second RICE table gives the equilibrium concentration for #"H"_3"O"^(+)#:

#["H"_3"O"^(+)] = color(purple)(x + y) ~~ 0.0803 color(white)(l) "mol" * "dm"^(-3)#

Jeremiah Adams · Answer

undefined To calculate the equilibrium concentration of H3O+ in a solution of oxalic acid, you need to consider the dissociation of oxalic acid. Oxalic acid (H2C2O4) is a weak acid that dissociates partially in water. Its dissociation equation is:

H2C2O4 (aq) ⇌ H+ (aq) + HC2O4- (aq)

Since oxalic acid is a diprotic acid, it dissociates in two steps:

1. H2C2O4 ⇌ H+ + HC2O4-
2. HC2O4- ⇌ H+ + C2O4^2-

Given that oxalic acid is a weak acid, you can use the initial concentration (0.20 M) to calculate the equilibrium concentration of H3O+ using the acid dissociation constant (Ka) of the first dissociation step (Ka1) for oxalic acid. The equilibrium concentrations can be found using an ICE table (Initial, Change, Equilibrium).

Ka1 for oxalic acid is typically provided in reference tables or can be calculated if the concentrations of the ions at equilibrium are known.

Let's denote:
[H2C2O4] = initial concentration of oxalic acid (0.20 M)
[H+] = concentration of H3O+ ions at equilibrium (unknown)
[HC2O4-] = concentration of the conjugate base at equilibrium (unknown)

Using the Ka1 expression:

Ka1 = [H+][HC2O4-] / [H2C2O4]

Since oxalic acid is a weak acid, you can assume that [H+] ≈ [HC2O4-], so you can simplify the expression:

Ka1 ≈ [H+]^2 / [H2C2O4]

Now, solve for [H+]:

[H+]^2 = Ka1 * [H2C2O4]

[H+] = √(Ka1 * [H2C2O4])

Plug in the values:

[H+] = √(Ka1 * 0.20 M)

Calculate the value using the given Ka1 and the initial concentration of oxalic acid.